$\begin{align}  & \angle XOY=2\times \angle XZY\text{  }\{\angle \text{ }at\text{ }centre=2\times \angle \text{ }at\text{ }circumference\} \\ & \angle XOY={{90}^{o}} \\ & \text{Area of }\vartriangle \text{XOY}=\frac{1}{2}{{r}^{2}}\sin \theta =\frac{1}{2}\times {{7}^{2}}\sin {{90}^{o}}=\frac{49}{2}c{{m}^{2}} \\ & \text{Area of sector }XOY=\frac{\theta }{{{360}^{o}}}\times \pi {{r}^{2}}=\frac{{{90}^{o}}}{{{360}^{o}}}\times \frac{22}{7}\times {{7}^{2}}=\frac{77}{2} \\ & \text{Area of the shaded portion}=\frac{77}{2}c{{m}^{2}}-\frac{49}{2}c{{m}^{2}} \\ & \text{Area of the shaded portion}=\frac{28}{2}c{{m}^{2}}=14c{{m}^{2}} \\\end{align}$